Yasmin-Leaves #31
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JS AdagramsWhat We're Looking For
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dHelmgren
reviewed
Nov 21, 2019
Comment on lines
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| let randomLetter = avaLetters[index]; | ||
| // remove the randomLetter from avaLetters array | ||
| avaLetters.splice(index, 1); | ||
| handOfLetters.push(randomLetter); |
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splice returns the deleted elements, so you can smoosh these lines together like so:
handOfLetters.push( avaLetters.splice(index, 1) );
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| const onePointLetter = ['A', 'E', 'I', 'O', 'U', 'L', 'N', 'R', 'S', 'T']; | ||
| const twoPointLetter = ['D', 'G']; | ||
| const threePointLetter = ['B', 'C', 'M', 'P']; | ||
| const fourPointLetter = ['F', 'H', 'V', 'W', 'Y']; | ||
| const fivePointLetter = ['K']; | ||
| const eightPointLetter = ['J', 'X']; | ||
| const tenPointLetter = ['Q', 'Z']; | ||
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| let letters = word.toUpperCase().split(''); | ||
| const scores = letters.map((letter) => { | ||
| let letterScore = 0; | ||
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| if (onePointLetter.includes(letter)) { | ||
| letterScore += 1; | ||
| } else if (twoPointLetter.includes(letter)) { | ||
| letterScore += 2; | ||
| } else if (threePointLetter.includes(letter)) { | ||
| letterScore += 3; | ||
| } else if (fourPointLetter.includes(letter)) { | ||
| letterScore += 4; | ||
| } else if (fivePointLetter.includes(letter)) { | ||
| letterScore += 5; |
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This ends up being pretty difficult to read and check. I would suggest using a JS Object in the form of
hashStyleScores = {
'A': 1,
'B': 4
}This format makes it easy to find and change each letter, and means that scoring a single letter is a one-line operation!
beccaelenzil
added a commit
to beccaelenzil/js-adagrams
that referenced
this pull request
Jun 2, 2022
remove references to Adagrams object
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JS Adagrams
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